Probability questions for various competitive examinations
In most of the competitive exams like
bank,IBPS,UPSC,SSC,MAT,CAT etc.., under quantitative section few questions are
asked from probability section. Here we are sharing important formulas and
facts,easy way to solve probability questions for competitive examinations. .We have tried to include all important formulas of probability. At the end you can also
find some solved practice questions on probability .
Prerequisites For probability section preparation-Knowledge of basics of set theory-permutation and combination
Probability-Important formulas and facts
Experiment:An operation which can produce some well-defined
outcomes is called an experiment.
Random Experiment:An experiment in which all possible
outcomes are known and exact output cannot be predicted in advance is called a
random experiments.For example, rolling a dice,tossing a coin,drawing a card
from a well shuffled pack of cards etc.
Sample space: When we perform an experiment, then the set of all possible outcomes is called Sample Space.denoted by ‘S’.
Event:Any subset of Sample Space is called an Event
Probability of occurrence of an Event:Let S be the sample space and E be the Event,then probability of occurrence of E denote by P(E)
P(E)=n(E)/n(S)=Number of favourable outcomes/Number of possible outcomes
or
P(E)=n(E)/n(S)=No of Trials in which E has happened / Total no of Trials
Some random experiments and their outcomes
Tossing a coin
When we toss a coin ,either Head(H) or a Tail(T) appears. If
two coins are tossed simultaneously then possible outcomes are HH,HT,TH and
TT.As number of coins increases possible outcomes also increases.
Rolling a Dice
A dice is a solid cube,having 6 faces,marked 1,2,3,4,5 and
6.When we roll a dice possible outcomes are 1,2,3,4,5 and 6.
If we roll two dice simultaneously posiible out comes are
combination of two number (1,1)(1,2)(1,3)……………….(6,6)
Card is drawn from a pack of cards
A pack of card has 52 cards.
It has 13 cards of each suit, namely Spades, Clubs, Hearts
and Diamonds.
Cards of Spade and Club are black.
Cards of Heart and Diamond are red cards.
There are 4 honours of each suit.These are Aces,Kings,Queens
and Jack.These are called face cards.
Elementary Event
An event having only one out come. Sum of probabilities of all the elementary events of an experiment is 1.
Eg. Rolling a dice, the outcome of getting 2 is an elementary event. Similarly outcome of getting 1,3,4,5 or 6is also elementary events.
P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1/6 +1/6 +1/6 +1/6 +1/6 +1/6 = 6/6 =1
Complementary Events
For any event E P(E)+P(E')=1, E and E' are complementary events
here E' is read as 'E not'
Example:Probability of getting an 'ace' from a pack of 52 cards
Let E be the event of getting an 'ace'
Since there are 4 aces in a pack P(E)=4/ 52
Probability of not getting an ace P(E')= 48/52
Here E and E' are complementary events.
P(E)+P(E')=1
Example:Tossing a coin , P(T)=1/2 , P(H)=1/2 and P(T)+P(H)=1
Here H & T are complementary events.
Mutually Exclusive Events(Disjoint Events)
Events That do not occur at the same time.
ie A ⋂ B=∅
Example:Tossing a coin, Head and Tail do not simultaneously.
Example:Rolling a die, 1 and 2 do not occur simultaneously.
Mutually Exclusice and Exhaustive events :
Events E1,E2,E3......En are mutually exclusive if Ei ⋂ Ej =∅ , where i≠j
and Exhaustive if
E1∪E2∪E3..........∪En=S
Conditional Probability
Conditional probability is the possibility of an event or outcome happening, based on the existence of a previous event or outcome.The conditional probability of an event E given the occurrence of the event F is given by
P(E|F)
P(E|F)=P(E ⋂ F) / P(F)
Example: Ten cards numbered 1 to 10 placed in a box and mixed up thoroughly and then one card is drawn randomly. If it is known that number on the card is more that 3 what is the probability that it is an even number?
Ans: Let A be the event 'the number on the card is even' and B be the event ' the number on the card is more than 3'
The sample space or all possible outcomes
S={1,2,3,4,5,6,7,8,9,10}
A={2,4,6,8,10}
B={4,5,6,7,8,9,10}
Conventional Method:
Here favourable out comes are {4,6,8,10}
Since it is given that card drawn is more than 3, new sample space or all possible outcomes =B={4,5,6,7,8,9,10}
Required probability P(A|B)=(No of favourable outcomes/ No of Total outcomes) = 4/7
Using Formula P(A|B)=P(A ⋂ B) / P(B)
A ⋂ B={4,6,8,10}
P(A ⋂ B)=4/10
P(B)=7/10
Now P(A|B)=P(A ⋂ B) / P(B) = (4/10)/ (7/10)=4/7
Results on probability
- P(S)=1 (Probability of sample space)
- Probability of a sure event is 1
- Probability of an impossible event is "0"
- 0≤P(E)≤1
- For any events A and B
- P(A)=1-P(not A)
Probability Practice questions
Question1:Two unbiased coin are tossed .What is the
probability of getting at most one head?
Solution: Sample space (All [possible
outcomes])S=(HH,HT,TH,TT)
Event(required outcomes)=(TT,HT,TH)
P(E)=n(E)/n(S)=3/4
Question2:Two dice are thrown simultaneously ,what is the
probability of getting a total of 7?
Solutions:n(S)=6*6=36
E={(1,6),(6,1),(5,2),(2,5),(4,3),(3,4)}
n(E)=6
P(E)=6/36=1/6
Question3:A bag contains 6 white and 4 black balls. Two
balls are drawn at random. Find the probability that they are of same colour.
Solution:n(S)=Number of ways two balls can be drawn from 10
balls(6 white +4 blacks)= 10C2=(10*9)/(2*1)=45
n(E)=Number of ways of drawing 2 balls from 6 white balls or
2 balls from 4 black balls=6C2+4C2=21
P(E)=21/45=7/15
Question4:Two cards are drawn at random from a pack of 52
cards.What is the probability that either both are black or both are queens?
Solution:Total number of possible ways selecting 2 card from
52 cards is n(S)=52C2=(52*51)/(2*1)=1326
Let A be event of getting both black card.
B be event of getting both queens.
Event of getting two black queens AB
There are 26 black cards in a pack and 4
queens. Two queens are black
n(A)=Number of ways of selecting 2 blacks from 26 cards=26C2=325
n(A∩B)=Chances of getting
2 black queens=2C2=1
P(A)=325/1326
P(B)=6/1326
P(A∩B)=1/1326
P(A ∪ B)=P(A)+P(B)-P(A∩B)
=(325/1326)+(6/1326)-(1/1326)
= 330/1326 =55/221
Tip:Use union U for ‘or’,
‘either’,’neither’
Use intersection ∩ for ‘and’.
If in the above question ,question if
changed like ‘chances of getting black queens’, then solution would be as
follows.(ie getting two card that are back and queens)
= P(A)+P(B)- P(A=1/1326
Question5:A speaks truth in 75% cases and B in 80% of cases. In what percentage of cases they
contradict each other in narrating the same incident?
Probability that A is telling the truth P(A)=75/100=3/4
Probability that B is telling the truth P(B)=80/100=4/5
Probability that A is lying P(A’)=1-75/100 =25/100=1/4
Probability that B is lying P(B’)=1-80/100 =20/100=1/5
Contradiction means either of
A telling
truth and B lying
Or
B telling truth
and A lying
P(Contradiction)= P(A)*P(B’) + P(A’)*P(B)=(3/4) * (1/5) +
(1/4)*(4/5)=7/20
7/20 is equal to 35 %.
Question6:A bag contains 2 red, 3 green and 2 blue balls.Two balls are drawn
at random.What is the probability that none of the drawn ball is blue?
Solution:Possible out comes(E) or Favourable outcomes are
E1.2 red ball
or
E2.2 green ball
or
E3.1 red and 1 green
Number of selecting 2 balls from two red balls n(E)=2C2=1
Number of ways of selecting 2 balls from 3 green balls
n(E2)= 3C2=3
Number of ways of selecting 1 red ball and 1 green ball
n(E3)= 2C1*3C1=2*3=6
Sample space S=Number of ways of selecting 2 balls from
7(2+3+2) balls n(S)=7C2=21
n(E)=n(E1)+ n(E2)+ n(E3)=1+3+6=10
Required probability P(E)=n(E)/n(S)=10/21
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